Question

In order to estimate the mean amount of time computer users spend on the internet each​ month, how many computer users must be surveyed in order to be 90​% confident that your sample mean is within 13 minutes of the population​ mean? Assume that the standard deviation of the population of monthly time spent on the internet is 228 min

Answer 1

Answer:

832

Step-by-step explanation:

standard deviation =228 minute

error =13 minute given

confidence level =905% =0.90

α=1-0.90=0.1

$z_\frac{\alpha }{2}=z_\frac{0.1}{2}=1.645$

we know that sample size should be greater than

$n\geq \left ( z_\frac{\alpha }{2}\times \frac{\sigma }{E} \right )^2$

$n\geq \left ( 1.645\times \frac{228}{13} \right )^{2}$

$n\geq 28.850^2$

$n\geq 832.3668$

n=832